Question

Suppose that each of five machines in a given shop breaks down according to a Poisson law at an average rate of one every 10 h, and the failures are repaired one at a time by two maintenance people operating as two channels, such that each machine has an exponentially distributed servicing requirement of mean 5 h.

(a) What is the probability that exactly one machine will be up at anyone time?

(b) If performance of the workforce is measured by the ratio of average waiting time to average service time, what is this measure for the current situation?

(c) What is the answer to (a) if an identical spare machine is put on reserve?

Answer #1

Now there are 5 machines and it breaks at an average rate of 1 every 10 hours, mean service time = 5 hrs

a. The poisson distribution has the follwing pdf f(x) = (e^-u) * (u^x) / x!, where f(x) is the probability of x successes in a given interval of time, e = 2.718 and u is the mean number of occurrences in the given interval therefore probability of exactly one machine being up is equal to the probability of 4 machines failing in the given interval of time

i.e. f(x) = (e^-1) * (1^4) / 4! = **0.015**

b. If measure of performance is ratio of average waiting time to average service time then

current measure = 5 hrs /10 hrs = **50%**

c. If an identical spare machine is put on reserve then we have
to measure the probability that all five fail in the 10 hr interval
to compute (a) = (e^-1) * 1^5 / 5! = **0.003**

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